Theorem ddif 4087

Description: Double complement under universal class. Exercise 4.10(s) of [Mendelson] p. 231. (Contributed by NM, 8-Jan-2002.)

Assertion

Ref	Expression
ddif	⊢ (V ∖ (V ∖ 𝐴)) = 𝐴

Proof of Theorem ddif

Dummy variable 𝑥 is distinct from all other variables.

Step	Hyp	Ref	Expression
1		velcomp 3913	. . . 4 ⊢ (𝑥 ∈ (V ∖ 𝐴) ↔ ¬ 𝑥 ∈ 𝐴)
2	1	con2bii 360	. . 3 ⊢ (𝑥 ∈ 𝐴 ↔ ¬ 𝑥 ∈ (V ∖ 𝐴))
3		vex 3454	. . . 4 ⊢ 𝑥 ∈ V
4	3	biantrur 540	. . 3 ⊢ (¬ 𝑥 ∈ (V ∖ 𝐴) ↔ (𝑥 ∈ V ∧ ¬ 𝑥 ∈ (V ∖ 𝐴)))
5	2, 4	bitr2i 279	. 2 ⊢ ((𝑥 ∈ V ∧ ¬ 𝑥 ∈ (V ∖ 𝐴)) ↔ 𝑥 ∈ 𝐴)
6	5	difeqri 4075	1 ⊢ (V ∖ (V ∖ 𝐴)) = 𝐴

This proof depends on syntax axioms:  ¬ wn 3   ∧ wa 401   = wceq 1570   ∈ wcel 2145  Vcvv 3450   ∖ cdif 3895

This proof depends on axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1828  ax-4 1842  ax-5 1943  ax-6 2000  ax-7 2041  ax-8 2147  ax-9 2155  ax-ext 2732

This proof depends on definitions:  df-bi 210  df-an 402  df-tru 1573  df-ex 1813  df-sb 2100  df-clab 2739  df-cleq 2752  df-clel 2835  df-v 3452  df-dif 3901

This theorem is used by:  complss  4097  dfun3  4221  dfin3  4222  invdif  4224  ssindif0  4416  difdifdir  4446